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How do you do these problems? Can you explain them fully.

1. Sin A Cot A =_______
2. Sec A Cos^2 A = _______
3.+ or - squareroot of 1+Tan^2 A = ________
4. __________ = + or - squareroot of cot^2 A+1
5. Csc A Tan A = _______
6. Sin A Sec A = ________
7.Cos A / Cot A = ______ (Is this Sin A?)
8. Cot A Sec A = _____

thanks!

2007-02-27 19:44:08 · 8 answers · asked by wordpuzzler_01 1 in Science & Mathematics Mathematics

8 answers

1: CotA expands as sinA/CosA
so sinA cotA = sinA*sinA/cosA=sin^2A/cosA

2: secA is the reciprocal of cosA
so SecA Cos^2A = (1/CosA)*(cos^2A)
= CosA

3: Rule is : 1+tan^2A= sec^2A
so + or - root(1+tan^2A)= + or - root(sec^2A)=
= + or - secA

4:Rule is 1+cot^2A = Cosec^ 2A
so root(1+cot^2A)= root(cosec^2A)= cosecA

5: Rule: cosecA= (1/sinA)
and tanA= SInA/cosA

so cosecAtanA= (1/sinA)*(sinA/CosA)= (1/cosA) = SecA

6: sinA(1/cosA) = sinA/cosA = tanA
7: cosA/cotA= cosA/(cosA/sinA)=sinA
8: (cosA/SinA)*(1/cosA)= (1/sinA)=cosecA

2007-02-27 20:02:58 · answer #1 · answered by vaidehi 2 · 0 0

firstly, tan A = sin A/cos A, and since cot A = 1/tan A, hence cot A = cos A/sin A. also, sec A = 1/cos A, and cosec A = 1/sin A.
so,
1. (sin A)(cot A) = (sin A)(cos A/sin A). note that sin A/sin A = 1. so the answer is cos A.
2. (sec A)(cos^2 A) = (1/cos A)(cos A x cos A)
the answer is easily cos A

one of the trigonometry rule is: sin^2 A + cos^2 A = 1.
so if you divide the entire equation with cos^2 A, you derive another rule that holds the same relationship: tan^2 A + 1 = sec^2 A.
on the other hand, if you divide the equation with sin^2 A instead of cos^2 A, you'll get another derivation! it becomes 1 + cot^2 A = cosec^2 A.
thus,
3. 1 + tan^2 A = sec^2 A. the +/- squareroot is thus just +/-(sec A)
4. cot^2 A + 1 = cosec^2 A. so, similarly, the +/- squareroot is +/-(cosec A)

5. cosec A tan A = (1/sin A)(sin A/cos A)
answer is 1/cos A, which is efectively sec A
6. sin A sec A = (sin A)(1/cos A) = sin A/cos A = tan A
7. cos A/cot A = cos A x tan A = (cos A)(sin A/cos A)
answer is, yes sin A.
8. cot A sec A = (cos A/sin A)(1/cos A) = 1/sin A = cosec A.

TIP: just convert everything to the most basic form, ie sin A and cos A only. then it'll be easier. at the same time, know what rules there are, and spot the pattern within the equations that you have to solve. trigonometry at this stage is very fun and easy.

2007-02-28 04:10:25 · answer #2 · answered by Anonymous · 0 0

Just remember that
cot = 1/tan
tan = sin / cos
csc = 1/sin
sec = 1/sin

Your question 7 is right. Cos / (1/Tan) = Cos * Tan = Cost *sin / cos = sin . You take a few different methods to figure otu this one, but they all end up the same.

Just do that for all of them.

Didn''t know what 1+tan^2 was so i looked it up
= sec^2...so +/- (sec A)

2007-02-28 04:18:56 · answer #3 · answered by My name is not bruce 7 · 0 0

convert all trigonometric function to f(sin A, cos A)ie
example tan A =sin A / cos A , cot A=cos A /sin A,
sec A =1/cos A , cosec A =1/sin A ;
USE 2ND CONCEPT (cosA)^2+(sin A)^2=1; You can find all solution using above procedures.

2007-02-28 04:09:51 · answer #4 · answered by rgfmss 2 · 0 0

hey.. thats too long a question ..! To start with

Sin A * Cot A = Sin A * (Cos A / Sin A )
{ Cot A is replaced by Cos A/Sin A)

= Cos A

use similar logic for all.

2007-02-28 03:51:29 · answer #5 · answered by HariSeldon 2 · 1 0

CotA = CosA/SinA
SecA = 1/CosA
CscA = 1/SinA

2007-02-28 04:00:29 · answer #6 · answered by Helmut 7 · 0 0

1 sinA(cosA/sinA)=cosA(coA=cosA/sinA)
2 cosA
3 +or-secA
4 +or-cosecA
5 secA
6 tanA
7 sinA
8 cosecA
check out ur text book for formulae and identities

2007-02-28 04:04:37 · answer #7 · answered by phani 1 · 0 0

I dont know,mathematics sucks!!

2007-02-28 03:51:37 · answer #8 · answered by Cadpigfriend 2 · 0 1

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